Consider the error integral
$$ I(x)=\int_{x}^{\infty} e^{-t^{2}} d t $$
for small $x$. Expanding the integrand in a Taylor series, we have
$$ e^{-t^{2}}=\sum_{n=0}^{\infty} \frac{(-1)^{n} t^{2 n}}{n !} $$
which converges for all values of $t$ as shown in the preceding example. However, $\textbf{no finite number of terms can represent $\mathbf{e^{-t^{2}}}$ for all t}!$
So we represent the error integral in the following alternate form:
$$ I(x)=\int_{0}^{\infty} e^{-t^{2}} d t-\int_{0}^{x} e^{-t^{2}} d t=\frac{\sqrt{\pi}}{2}-\int_{0}^{x}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !} t^{2 n} d t $$
The above is from my textbook and I'm confused about what the bold part exactly says. I know substituting the Taylor series into the original integral does not work unless doing the transformation but what does the bold typeface mean?
The statement in bold reflects the fact that as $t \to \infty$ the terms in the series get very large in magnitude, so that if you truncate the series, the term at the end gets too big. The series itself converges for each $t$, since the $n!$ eventually dominates $t^n$. Truncation is not a problem for the second expression, since there is an upper limit for $t$, namely $x$.